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Let $a_0 := 1$ and $b_0 := \sqrt{2}$, and define \begin{align*} a_{i+1} &:= |b_i - a_i| \\ b_{i+1} &:= |a_i - a_{i+1}| \end{align*} Prove that $\lim_{i \to \infty}{a_i} = 0$ and $\lim_{i \to \infty}{b_i} = 0$.

Context: I am given the set $L := \{a+b\sqrt{2} : a,b\in \mathbb{Z}\} \subset \mathbb{R}$, and I want to prove it is NOT a lattice over $\mathbb{R}$, since it is not discrete. My idea to solve it was to build a sequence converging to $0$ and I came up with this sequence. Intuitively/empirically seems to be converging to $0$, but I am having trouble proving it formally. Does anyone have any idea or hint?

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    $\begingroup$ You probably don't need to solve this problem about sequences. The statement that you are trying to prove in the first place ($L$ not discrete) is the consequence of Dirichlet's approximation theorem. $\endgroup$ May 20 at 8:24
  • $\begingroup$ @StinkingBishop Uh, indeed, thanks! But honestly now I am still curious to see how to prove the convergence of this sequence. $\endgroup$
    – Ronteg
    May 20 at 8:35
  • $\begingroup$ @StinkingBishop By the way, the application of the Dirichlet's approximation would still not be so straightforward, because I cannot build rational numbers in the lattice (as additive subgroup of $\mathbb{R}$). $\endgroup$
    – Ronteg
    May 20 at 8:39
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    $\begingroup$ Well, take $\alpha=\sqrt{2}$ (the symbols from Wikipedia) and for every $N$ you can find $p,q$ such. that $|q\sqrt{2}-p|<\frac{1}{N}$. Thus, you can find numbers of the form $a+b\sqrt{2}$ arbitrarily close to zero. (Take $a=-p, b=q$.) $\endgroup$ May 20 at 8:43
  • $\begingroup$ @StinkingBishop Uh, I was focusing too much on the "any real number has a sequence of good rational approximations" part. Thanks for the explicit answer! $\endgroup$
    – Ronteg
    May 20 at 8:49

2 Answers 2

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Prove, by induction that

$$a_n = \left(\sqrt{2} - 1\right)^n, ~~b_n = \left(\sqrt{2} - 1\right)^{n-1} - \left(\sqrt{2} - 1\right)^n. \tag1 $$


Empirically true for $n=1$.

$$a_{n+1} = |b_n - a_n| = |\left(\sqrt{2} - 1\right)^{n-1} \times [ 1 - 2(\sqrt{2} - 1) ]|$$

$$ = |\left(\sqrt{2} - 1\right)^{n-1} \times (3 - 2\sqrt{2})|$$

$$ = |\left(\sqrt{2} - 1\right)^{n-1} \times \left(\sqrt{2} - 1\right)^2| = \left(\sqrt{2} - 1\right)^{n+1}.$$

Therefore,

$$b_{n+1} = |a_{n+1} - a_{n}| = \left(\sqrt{2} - 1\right)^n - \left(\sqrt{2} - 1\right)^{n+1}.$$

So, (1) above is proven.

Further, it is immediate from (1) above that as $n \to \infty, a_n \to 0.$

Finally, $b_n$ may be re-written

$$\left(\sqrt{2 - 1}\right)^{n-1} [1 - (\sqrt{2} - 1)] = (2-\sqrt{2})a_{n-1}.$$

So, as $n \to \infty, b_n \to 0.$

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  • $\begingroup$ Amazing, thanks! Can I ask you how did you come up with the induction thesis in the first place? Just by intuition? $\endgroup$
    – Ronteg
    May 20 at 9:39
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    $\begingroup$ @Ronteg No, by elbow grease. I manually charted $a_n,b_n$ for $n \in \{0,1,2,3,4\}$, and noticed how, for $a_n$, the expressions changed from $R - S\sqrt{2}$ to $T\sqrt{2} - U$ to $V - W\sqrt{2}$. Then, I just sat and meditated. $\endgroup$ May 20 at 9:41
  • $\begingroup$ Back to the motivation, you only need $(\sqrt{2}-1)^n\rightarrow 0$. $\endgroup$ May 20 at 9:47
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    $\begingroup$ @Justauser My intuition isn't that strong, and my knowledge of Real Analysis is shallow. For example, I have never heard of Dirichlet. So, I needed to see the values, and think about them. Hence, the elbow grease. $\endgroup$ May 20 at 9:50
  • $\begingroup$ @user2661923 what you did is amazing, as people often say mathematics might very well be an experimental subject in practice. I just want to say to Ronteg that we don't really need $b_n$ once this clears. $\endgroup$ May 20 at 9:56
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If all you want is to show $L$ is discrete, it's much simpler than Dirichlet's theorem: the sequence $\{x_n=a_n+b_n\sqrt{2}| b_n = n, a=-\lfloor b\sqrt 2\rfloor\}_{n=0}^\infty$ is an infinite bounded subset of $[0, 1)$. So $L$ cannot be discrete. To be more precise, it contains a convergent subsequence $\{x_{n_i}\}$, then $|x_{n_i}-x_{n_j}|>0$ can be arbitrarily small for suitable $i,j$.

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    $\begingroup$ Thanks for the answer this is quite straightforward indeed. But I still hope someone can help me out with that sequence, otherwise I will keep thinking about it for weeks. $\endgroup$
    – Ronteg
    May 20 at 8:54

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